The value of four quantum numbers for the last electron of atom of element 'P' are $\mathrm{n}=7,\mathrm{l}=1,\mathrm{m}=\pm 1\text{ or }\mathrm{s}=+\frac{1}{2},-\frac{1}{2}$ and value of spin magnetic moment for element 'P' is zero. The element has two isotopes, ${ }_z^{\mathrm{A}}\mathrm{P}{\text{ and }}_z^{\mathrm{B}}\mathrm{P}$. 

Given: B - A = B - 2Z = 18, where A and B are atomic masses and Z is atomic number.
Select incorrect statement based upon above information 

$\text{ For last }{e}^{-}\text{of element ' }{P}^{\mathrm{\prime }}\Rightarrow n=7,l=1,m=+1\text{ and }s=$ $+\frac{1}{2}\text{ or }-\frac{1}{2}\text{, shows that last }{e}^{-}\text{is present in " }7p\text{ " sub shell and }$spin magnetic moment is zero, so there is no unpaired e^-1s.
Hence outer electronic configuration of "P' - 7P⁶ ; belongs to 18th group with Z = 118.
Correct (2)
'P' is nobel gas and is not representative element according to Bohr's classification.

$\text{ (3) }Z=118,B-2Z=18\text{, so }B=254\text{ and }B-A=18\text{, so }A=236\text{. }$

$\text{ (4) }90\text{ th }{e}^{-}\text{goes to }5f\text{ subshell so four quantum numbers are, }$

$\mathrm{n}=5,\mathrm{l}=3,\mathrm{m}=-2\text{ and }\mathrm{s}=-\frac{1}{2}\text{ or }+\frac{1}{2}$